Shell script help for listing files

Hi,

I need to work on script which will give me the selective list of files in a directory into another file. I will explain , I have a script abc.sh , the shell is ksh. I have a directory /home/anand/output. The output will have files by the name file1.txt , file2.txt and file1.txt.1 and file2.txt.2. Now I want to get a list of files without the .1 and .2 extensions only. and put them into a list.txt file. So my list.txt will have file1.txtand file2.txt. Can someone help me with this. I am not being able to figure out how to not select files with the last alphabet as a numeral.

Oh no!! I just figured that the files might end with CXX. Which means C01, C02. so just giving "txt" will not help. I need something with egrep.

0

anandabrataAuthor Commented: 2007-03-22

Hi,

This command ( ls -l | egrep -v '\.[0-9][0-9]*$') works great excepting that it will also list the directories in which it runs. I have another saving grace i.e all the files that I want will always start with 2112657. So in addition in the above condition , even this needs to be checked. I tried
ls | egrep -v '\.[0-9][0-9]*$' | ls | grep '2112657*' and
ls | egrep -v '\.[0-9][0-9]*$' > ls | grep '2112657*'
but they don't work.

Thanks that helps a lot. I just need help with another thing, I want make a recursive call to the ls command used above. Like this
acty_loa_cnt=do a select from database which will retrun me a count value
if [ $acty_loa_cnt -eq 11 ]
then
ls | egrep -v '\.[0-9][0-9]*$' | grep '^211'
else
sleep 1000
fi
#do the select and if condition block above again. Basically make it recursive with a sleep of 1 min. how is that possible. Btw appreciate your help.